Using Lambda Expressions with remove_if() in std::forward_list

The remove_if() member function in std::forward_list accepts a unary predicate that returns true for the elements to be removed. We can use a lambda expression to define this predicate inline.

Here's an example that removes all even numbers from a forward_list:

1#include <forward_list>
2#include <iostream>
3
4int main() {
5  std::forward_list<int> list{
6    1, 2, 3, 4, 5, 6};
7
8  list.remove_if([](int i) {
9    return i % 2 == 0; });  
10
11  for (int i : list) {
12    std::cout << i << ' ';
13  }
14}

11 3 5

The lambda expression [](int i) { return i % 2 == 0; } returns true for even numbers, so remove_if() removes all even numbers from the list.

We can also capture variables in the lambda:

1#include <forward_list>
2#include <iostream>
3
4int main() {
5  std::forward_list<int> list{1, 2, 3, 4, 5, 6};
6  int threshold = 3;
7
8  list.remove_if([threshold](int i) {
9    return i > threshold; });  
10
11  for (int i : list) {
12    std::cout << i << ' ';
13  }
14}

11 2 3

Here, the lambda captures threshold by value and removes all elements greater than threshold.

Lambda expressions provide a convenient way to define simple predicates directly at the call site, improving readability and reducing the need for separate named functions.